You have found the following ages (in years) of all 5 turtles at your local zoo: $ 92,\enspace 57,\enspace 13,\enspace 107,\enspace 51$ What is the average age of the turtles at your zoo? What is the standard deviation? You may round your answers to the nearest tenth.
Explanation: Because we have data for all 5 turtles at the zoo, we are able to calculate the population mean $({\mu})$ and population standard deviation $({\sigma})$ To find the population mean , add up the values of all $5$ ages and divide by $5$ $ {\mu} = \dfrac{\sum\limits_{i=1}^{{N}} x_i}{{N}} = \dfrac{\sum\limits_{i=1}^{{5}} x_i}{{5}} $ $ {\mu} = \dfrac{92 + 57 + 13 + 107 + 51}{{5}} = {64\text{ years old}} $ Find the squared deviations from the mean for each turtle. Age $x_i$ Distance from the mean $(x_i - {\mu})$ $(x_i - {\mu})^2$ $92$ years $28$ years $784$ years $^2$ $57$ years $-7$ years $49$ years $^2$ $13$ years $-51$ years $2601$ years $^2$ $107$ years $43$ years $1849$ years $^2$ $51$ years $-13$ years $169$ years $^2$ Because we used the population mean $({\mu})$ to compute the squared deviations from the mean , we can find the variance $({\sigma^2})$ , without introducing any bias, by simply averaging the squared deviations from the mean $ {\sigma^2} = \dfrac{\sum\limits_{i=1}^{{N}} (x_i - {\mu})^2}{{N}} $ $ {\sigma^2} = \dfrac{{784} + {49} + {2601} + {1849} + {169}} {{5}} $ $ {\sigma^2} = \dfrac{{5452}}{{5}} = {1090.4\text{ years}^2} $ As you might guess from the notation, the population standard deviation $({\sigma})$ is found by taking the square root of the population variance $({\sigma^2})$ ${\sigma} = \sqrt{{\sigma^2}}$ $ {\sigma} = \sqrt{{1090.4\text{ years}^2}} = {33\text{ years}} $ The average turtle at the zoo is 64 years old. There is a standard deviation of 33 years.